# How do you find the critical numbers of  y= x^2*(1 + 3 ln x)?

Apr 2, 2018

$x = {e}^{- \frac{5}{6}}$

#### Explanation:

Differentiate, set equal to zero, and solve for $x ,$ in addition to being aware of specific values for which the derivative doesn't exist.

$y ' = {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left(1 + 3 \ln x\right) + \left(1 + 3 \ln x\right) \cdot \frac{d}{\mathrm{dx}} {x}^{2}$

$y ' = {x}^{2} \left(\frac{3}{x}\right) + 2 x \left(1 + 3 \ln x\right)$

$y ' = 3 x + 2 x + 6 x \ln x$

$y ' = 5 x + 6 x \ln x$

The derivative is logarithmic like the original function. They both share the domain $\left(0 , \infty\right) ,$ so there aren't any particular values for which the derivative doesn't exist while the function does.

$5 x + 6 x \ln x = 0$

$x \left(5 + 6 \ln x\right) = 0$

$x = 0$ looks like it would be a solution, but since it's not in the domain of the logarithm, it is not valid.

$5 + 6 \ln x = 0$

$6 \ln x = - 5$

$\ln x = - \frac{5}{6}$

${e}^{\ln} x = {e}^{- \frac{5}{6}}$

From the fact that ${e}^{\ln} x = x :$

$x = {e}^{- \frac{5}{6}}$

Is the only critical number.